Pips answers for Monday, November 17, 2025

Start with the “=” region at the top: all 3 cells must show the same number, and only 5s appear on enough dominoes to fill it.

• the “=” region at the top: 5, 5, 5
• the single-cell “<4” region at the bottom-left: 0
• the 4 region at the bottom: 1, 3
• the single-cell 4 region at the bottom-right: 4

1. The single-cell 4 region at the bottom-right needs a total of exactly 4 — nothing else fits. Place the 2–4 domino horizontally on row 3, column 4 and row 3, column 5.
2. The “=” region at the top needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino vertically on row 1, column 4 and row 2, column 4.
3. The “=” region at the top needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 5–6 domino horizontally on row 1, column 2 and row 1, column 3.
4. The 4 region at the bottom needs a total of exactly 4 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–1 domino horizontally on row 3, column 1 and row 3, column 2.
5. Row 3, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 4 region at the bottom needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 2, column 3 and row 3, column 3.

This board has more than one valid layout; the steps follow the official answer.

Start with the “=” region at the right: all 4 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the single-cell “>2” region at the top: 6
• the “=” region at the right: 1, 1, 1, 1
• the single-cell “>2” region at the top-right: 5
• the “=” region at the left: 4, 4, 4
• the “=” region at the center: 3, 3, 3
• the single-cell “>2” region at the bottom: 5

1. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the “=” region at the left needs every pip to match — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–4 domino horizontally on row 2, column 1 and row 2, column 2.
2. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 3, column 2 and row 3, column 3.
3. The “=” region at the right needs every pip to match — 5 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–5 domino horizontally on row 1, column 6 and row 1, column 7.
4. The “=” region at the right needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 1–1 domino vertically on row 2, column 6 and row 3, column 6.
5. The “=” region at the right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 0–1 domino vertically on row 1, column 5 and row 2, column 5.
6. The “=” region at the center needs every pip to match — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–5 domino horizontally on row 3, column 4 and row 3, column 5.
7. The “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–3 domino horizontally on row 2, column 3 and row 2, column 4.
8. Row 1, column 4 has only one open neighbor, so its domino must run to row 1, column 3; the single-cell “>2” region at the top must total more than 2 — nothing else fits. Place the 2–6 domino horizontally on row 1, column 3 and row 1, column 4.

Start with the 24 region at the top: the only way to make it with these dominoes is 6 + 6 + 6 + 6.

• the single-cell “>2” region at the top-left: 3
• the 24 region at the top: 6, 6, 6, 6
• the 10 region at the top-right: 5, 5
• the single-cell “>3” region at the top-left: 4
• the “≠” region at the center: 0, 1, 2, 3, 4, 5
• the single-cell 2 region at the center: 2
• the single-cell 0 region at the right: 0
• the single-cell 3 region at the center: 3
• the single-cell 1 region at the right: 1
• the single-cell 3 region at the bottom-right: 3
• the single-cell 5 region at the bottom: 5
• the single-cell 0 region at the bottom-left: 0
• the single-cell 3 region at the bottom: 3
• the single-cell 1 region at the bottom-right: 1

1. The 24 region at the top needs a total of exactly 24 — nothing else fits. Place the 4–6 domino horizontally on row 2, column 1 and row 2, column 2.
2. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the single-cell “>2” region at the top-left must total more than 2 — nothing else fits. Place the 3–6 domino horizontally on row 1, column 1 and row 1, column 2.
3. The 10 region at the top-right needs a total of exactly 10 — nothing else fits. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
4. The 24 region at the top needs a total of exactly 24 (2 of its 4 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–6 domino vertically on row 2, column 3 and row 3, column 3.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 24 region at the top needs a total of exactly 24 (3 of its 4 cells are already filled) — nothing else fits. Place the 1–6 domino vertically on row 2, column 4 and row 3, column 4.
6. The single-cell 2 region at the center needs a total of exactly 2 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–2 domino horizontally on row 4, column 2 and row 4, column 3.
7. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the single-cell 0 region at the right needs a total of exactly 0 — nothing else fits. Place the 0–1 domino vertically on row 4, column 4 and row 5, column 4.
8. The “≠” region at the center needs every pip to be different (2 of its 6 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino horizontally on row 5, column 2 and row 5, column 3.
9. The “≠” region at the center needs every pip to be different (3 of its 6 cells are already filled) — nothing else fits. Place the 3–3 domino horizontally on row 6, column 3 and row 6, column 4.
10. The single-cell 3 region at the bottom needs a total of exactly 3 — nothing else fits. Place the 0–3 domino horizontally on row 8, column 1 and row 8, column 2.
11. The “≠” region at the center needs every pip to be different (4 of its 6 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 8, column 3 and row 8, column 4.
12. The “≠” region at the center needs every pip to be different (5 of its 6 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 7, column 3 and row 7, column 4.
13. Row 7, column 2 has only one open neighbor, so its domino must run to row 7, column 1; the single-cell 5 region at the bottom needs a total of exactly 5 — nothing else fits. Place the 2–5 domino horizontally on row 7, column 1 and row 7, column 2.

This board has more than one valid layout; the steps follow the official answer.